Pirate democracy at its finestPirates and gold coinsThe coolest checkerboard magic trickThe Trickster's GamePirate Puzzle Double TwistIf 27 x 147 = 3969 What is its square root WITHOUT using complex / long calculationx
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Pirate democracy at its finest
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Pirate democracy at its finest
Pirates and gold coinsThe coolest checkerboard magic trickThe Trickster's GamePirate Puzzle Double TwistIf 27 x 147 = 3969 What is its square root WITHOUT using complex / long calculationx
$begingroup$
With our pirate crew becoming too big, the captain grew very concerned about splitting all the treasure - we continued to split it equally, but, of course, each crew member got less and less with the time.
When the captain got sick of it, he gathered all the crew and made an announcement:
Pirates! There are already 999 of you in my crew! That's way too much!
I gave each of us a different number from 1 to 1000 according to how
much you do for the crew. I myself got the 1000, and Michael, the one
sleeping in that corner, got the 1.
From now on we will each day vote on executing the lamest member of
our crew, in the order: 1, 2, 3, etc... The one who is judged
doesn't vote! If the strict majority (> 0.5) of others decides to execute the
lamest member, we do so, and then continue. If not - we stop the
process altogether.
That is a completely democratic way to clear the crew of the weakest
members. The ones alive will totally benefit from it, for their share
in the treasures we pillage will highly increase!
Given that every pirate was very clever and predictive, how many pirates died in the process?
strategy combinatorics
edited 8 hours ago
Glorfindel
15.6k45890
asked 8 hours ago
Thomas BlueThomas Blue
2,7041649
$endgroup$
add a comment |
$begingroup$
With our pirate crew becoming too big, the captain grew very concerned about splitting all the treasure - we continued to split it equally, but, of course, each crew member got less and less with the time.
When the captain got sick of it, he gathered all the crew and made an announcement:
Pirates! There are already 999 of you in my crew! That's way too much!
I gave each of us a different number from 1 to 1000 according to how
much you do for the crew. I myself got the 1000, and Michael, the one
sleeping in that corner, got the 1.
From now on we will each day vote on executing the lamest member of
our crew, in the order: 1, 2, 3, etc... The one who is judged
doesn't vote! If the strict majority (> 0.5) of others decides to execute the
lamest member, we do so, and then continue. If not - we stop the
process altogether.
That is a completely democratic way to clear the crew of the weakest
members. The ones alive will totally benefit from it, for their share
in the treasures we pillage will highly increase!
Given that every pirate was very clever and predictive, how many pirates died in the process?
strategy combinatorics
edited 8 hours ago
Glorfindel
15.6k45890
asked 8 hours ago
Thomas BlueThomas Blue
2,7041649
$endgroup$
add a comment |
$begingroup$
With our pirate crew becoming too big, the captain grew very concerned about splitting all the treasure - we continued to split it equally, but, of course, each crew member got less and less with the time.
When the captain got sick of it, he gathered all the crew and made an announcement:
Pirates! There are already 999 of you in my crew! That's way too much!
I gave each of us a different number from 1 to 1000 according to how
much you do for the crew. I myself got the 1000, and Michael, the one
sleeping in that corner, got the 1.
From now on we will each day vote on executing the lamest member of
our crew, in the order: 1, 2, 3, etc... The one who is judged
doesn't vote! If the strict majority (> 0.5) of others decides to execute the
lamest member, we do so, and then continue. If not - we stop the
process altogether.
That is a completely democratic way to clear the crew of the weakest
members. The ones alive will totally benefit from it, for their share
in the treasures we pillage will highly increase!
Given that every pirate was very clever and predictive, how many pirates died in the process?
strategy combinatorics
edited 8 hours ago
Glorfindel
15.6k45890
asked 8 hours ago
Thomas BlueThomas Blue
2,7041649
$endgroup$
With our pirate crew becoming too big, the captain grew very concerned about splitting all the treasure - we continued to split it equally, but, of course, each crew member got less and less with the time.
When the captain got sick of it, he gathered all the crew and made an announcement:
Pirates! There are already 999 of you in my crew! That's way too much!
I gave each of us a different number from 1 to 1000 according to how
much you do for the crew. I myself got the 1000, and Michael, the one
sleeping in that corner, got the 1.
From now on we will each day vote on executing the lamest member of
our crew, in the order: 1, 2, 3, etc... The one who is judged
doesn't vote! If the strict majority (> 0.5) of others decides to execute the
lamest member, we do so, and then continue. If not - we stop the
process altogether.
That is a completely democratic way to clear the crew of the weakest
members. The ones alive will totally benefit from it, for their share
in the treasures we pillage will highly increase!
Given that every pirate was very clever and predictive, how many pirates died in the process?
strategy combinatorics
strategy combinatorics
edited 8 hours ago
Glorfindel
15.6k45890
asked 8 hours ago
Thomas BlueThomas Blue
2,7041649
edited 8 hours ago
Glorfindel
15.6k45890
asked 8 hours ago
Thomas BlueThomas Blue
2,7041649
edited 8 hours ago
Glorfindel
15.6k45890
edited 8 hours ago
Glorfindel
15.6k45890
edited 8 hours ago
Glorfindel
15.6k45890
15.6k45890
asked 8 hours ago
Thomas BlueThomas Blue
2,7041649
asked 8 hours ago
Thomas BlueThomas Blue
2,7041649
asked 8 hours ago
Thomas BlueThomas Blue
2,7041649
2,7041649
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I have a hunch that the answer is
489, so 511 pirates remain.
Explanation:
When there is 1 pirate left, obviously nothing happens.
When there are 2 pirates left, the one with the higher number will vote to execute the other so he gets a larger share.
When there are 3 pirates left, the one with the second-highest number will not vote to execute, since that would leave him in the previous situation where he will be executed. Since the votes are tied, the process stops here.
When there are 4 pirates left, the three pirates will vote to execute the last one in order to get a larger share. There's no risk that they will be executed, since the process will stop at 3.
When there are 5 pirates left, the one with the fourth-highest number cannot stop the other three to reach the situation with 4 pirates. The same holds for 6 pirates.
When there are 7 pirates left, the three pirates 4-6 can vote not to execute in order to stop being executed themselves in the next steps. So 7 is again a 'stable' number.
Continuing this way, we see that
when there are $2^n-1$ pirates left, execution will stop.
Since 511 is the largest such number smaller than 1000, 511 pirates will remain alive and 489 will die.
answered 8 hours ago
GlorfindelGlorfindel
15.6k45890
$endgroup$
add a comment |
$begingroup$
This is the same exact problem (actually off-by one due to your voting rules) of Pirates and gold coins with 0 coins.
This is a great answer by SQB which explains the logic in detail: and gets the same result as Glorfindel's.
edited 4 mins ago
Rupert Morrish
3,7631935
answered 25 mins ago
12345ieee12345ieee
112
New contributor
12345ieee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I have a hunch that the answer is
489, so 511 pirates remain.
Explanation:
When there is 1 pirate left, obviously nothing happens.
When there are 2 pirates left, the one with the higher number will vote to execute the other so he gets a larger share.
When there are 3 pirates left, the one with the second-highest number will not vote to execute, since that would leave him in the previous situation where he will be executed. Since the votes are tied, the process stops here.
When there are 4 pirates left, the three pirates will vote to execute the last one in order to get a larger share. There's no risk that they will be executed, since the process will stop at 3.
When there are 5 pirates left, the one with the fourth-highest number cannot stop the other three to reach the situation with 4 pirates. The same holds for 6 pirates.
When there are 7 pirates left, the three pirates 4-6 can vote not to execute in order to stop being executed themselves in the next steps. So 7 is again a 'stable' number.
Continuing this way, we see that
when there are $2^n-1$ pirates left, execution will stop.
Since 511 is the largest such number smaller than 1000, 511 pirates will remain alive and 489 will die.
answered 8 hours ago
GlorfindelGlorfindel
15.6k45890
$endgroup$
add a comment |
$begingroup$
I have a hunch that the answer is
489, so 511 pirates remain.
Explanation:
When there is 1 pirate left, obviously nothing happens.
When there are 2 pirates left, the one with the higher number will vote to execute the other so he gets a larger share.
When there are 3 pirates left, the one with the second-highest number will not vote to execute, since that would leave him in the previous situation where he will be executed. Since the votes are tied, the process stops here.
When there are 4 pirates left, the three pirates will vote to execute the last one in order to get a larger share. There's no risk that they will be executed, since the process will stop at 3.
When there are 5 pirates left, the one with the fourth-highest number cannot stop the other three to reach the situation with 4 pirates. The same holds for 6 pirates.
When there are 7 pirates left, the three pirates 4-6 can vote not to execute in order to stop being executed themselves in the next steps. So 7 is again a 'stable' number.
Continuing this way, we see that
when there are $2^n-1$ pirates left, execution will stop.
Since 511 is the largest such number smaller than 1000, 511 pirates will remain alive and 489 will die.
answered 8 hours ago
GlorfindelGlorfindel
15.6k45890
$endgroup$
add a comment |
$begingroup$
I have a hunch that the answer is
489, so 511 pirates remain.
Explanation:
When there is 1 pirate left, obviously nothing happens.
When there are 2 pirates left, the one with the higher number will vote to execute the other so he gets a larger share.
When there are 3 pirates left, the one with the second-highest number will not vote to execute, since that would leave him in the previous situation where he will be executed. Since the votes are tied, the process stops here.
When there are 4 pirates left, the three pirates will vote to execute the last one in order to get a larger share. There's no risk that they will be executed, since the process will stop at 3.
When there are 5 pirates left, the one with the fourth-highest number cannot stop the other three to reach the situation with 4 pirates. The same holds for 6 pirates.
When there are 7 pirates left, the three pirates 4-6 can vote not to execute in order to stop being executed themselves in the next steps. So 7 is again a 'stable' number.
Continuing this way, we see that
when there are $2^n-1$ pirates left, execution will stop.
Since 511 is the largest such number smaller than 1000, 511 pirates will remain alive and 489 will die.
answered 8 hours ago
GlorfindelGlorfindel
15.6k45890
$endgroup$
I have a hunch that the answer is
489, so 511 pirates remain.
Explanation:
When there is 1 pirate left, obviously nothing happens.
When there are 2 pirates left, the one with the higher number will vote to execute the other so he gets a larger share.
When there are 3 pirates left, the one with the second-highest number will not vote to execute, since that would leave him in the previous situation where he will be executed. Since the votes are tied, the process stops here.
When there are 4 pirates left, the three pirates will vote to execute the last one in order to get a larger share. There's no risk that they will be executed, since the process will stop at 3.
When there are 5 pirates left, the one with the fourth-highest number cannot stop the other three to reach the situation with 4 pirates. The same holds for 6 pirates.
When there are 7 pirates left, the three pirates 4-6 can vote not to execute in order to stop being executed themselves in the next steps. So 7 is again a 'stable' number.
Continuing this way, we see that
when there are $2^n-1$ pirates left, execution will stop.
Since 511 is the largest such number smaller than 1000, 511 pirates will remain alive and 489 will die.
answered 8 hours ago
GlorfindelGlorfindel
15.6k45890
answered 8 hours ago
GlorfindelGlorfindel
15.6k45890
answered 8 hours ago
GlorfindelGlorfindel
15.6k45890
answered 8 hours ago
GlorfindelGlorfindel
15.6k45890
15.6k45890
add a comment |
add a comment |
$begingroup$
This is the same exact problem (actually off-by one due to your voting rules) of Pirates and gold coins with 0 coins.
This is a great answer by SQB which explains the logic in detail: and gets the same result as Glorfindel's.
edited 4 mins ago
Rupert Morrish
3,7631935
answered 25 mins ago
12345ieee12345ieee
112
New contributor
12345ieee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This is the same exact problem (actually off-by one due to your voting rules) of Pirates and gold coins with 0 coins.
This is a great answer by SQB which explains the logic in detail: and gets the same result as Glorfindel's.
edited 4 mins ago
Rupert Morrish
3,7631935
answered 25 mins ago
12345ieee12345ieee
112
New contributor
12345ieee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This is the same exact problem (actually off-by one due to your voting rules) of Pirates and gold coins with 0 coins.
This is a great answer by SQB which explains the logic in detail: and gets the same result as Glorfindel's.
edited 4 mins ago
Rupert Morrish
3,7631935
answered 25 mins ago
12345ieee12345ieee
112
New contributor
12345ieee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This is the same exact problem (actually off-by one due to your voting rules) of Pirates and gold coins with 0 coins.
This is a great answer by SQB which explains the logic in detail: and gets the same result as Glorfindel's.
edited 4 mins ago
Rupert Morrish
3,7631935
answered 25 mins ago
12345ieee12345ieee
112
New contributor
12345ieee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 mins ago
Rupert Morrish
3,7631935
edited 4 mins ago
Rupert Morrish
3,7631935
edited 4 mins ago
Rupert Morrish
3,7631935
3,7631935
answered 25 mins ago
12345ieee12345ieee
112
New contributor
12345ieee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 25 mins ago
12345ieee12345ieee
112
answered 25 mins ago
12345ieee12345ieee
112
112
New contributor
12345ieee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
12345ieee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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